Population and Percentage
Date: 03/07/99 at 23:12:30 From: Ian Hubble and Ross Hubble Subject: Population and Percentage Between 1902 and 1952, the population of Maranga increased by 30,000. Dividing the population up into adults and children, and males and females, the following information can be given about the number of men, women, boys, and girls. In 1902, the census showed that 60% of the people were males. There were half as many boys as females, and for every four boys there were three girls. At the time of the 1952 census, 25% of the total were children. The number of adults was five times the number of boys, and there were as many women as there were males. The number of children equalled the number of females in 1902. How many women were there in 1902, and how many in 1952? We have worked through this and by logic and algebra have found 1902 1952 men 40% 30% women 25% 45% boys 20% 15% girls 15% 10% What we are stuck on is linking the 30,000 increase and % change to work out the number of women. Can you please help?
Date: 03/08/99 at 05:37:58 From: Doctor Allan Subject: Re: Population and Percentage If we call the population number in 1902 X and the population number in 1952 Y, we have two useful pieces of information. We know that the number of children in 1952 equals the number of females in 1902. Since the number of females equals 40% of the total population in 1902 and the number of children equals 25% of the population in 1952, we have: (1) 0.40*X = 0.25*Y Furthermore, we know that the population increase is 30,000. This means that (2) Y - X = 30000 We now have two equations with two unknowns, which we can solve the usual way. Isolate Y in (2) to get (2') Y = 30000 + X and substitute this into (1) yielding (1') 0.40*X = 0.25*(30000 + X) 0.40*X = 7500 + 0.25*X 0.15*X = 7500 X = 50000 Substituting this value into (2') yields (2'') Y = 80000 We have figured out the total population of 1902 and 1952. Now the number of women can be easily calculated by using their percentage-rate of the total population. - Doctor Allan, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum